Category Archives: Tech

PP Fixed Bias circuit design and calculator

The push-pull design for output stages has persisted since times of antiquity. It was one of the very earliest circuit designs, and has persisted until the present day, with modern solid-state linear amplifiers still overwhelmingly using it.

With tubes, a typical topology is given by the circuit below. (Click to magnify). The anode (plate) voltage on either side comes through the primary of the output transformer.
This design uses a pentode tube, which has a screen grid. This is attached to taps on the output transformer to run in Ultra-Linear mode, increasing efficiency and reducing distortion.

In this design, the cathodes are tied to ground through a very low value shunt resistor. The resistor is simply there to provide a small voltage drop from which the current through the tube can be measured. It plays no other role in the circuit, other than being a fuse if the tube red-plates.

Values of one ohm or ten ohms are typical of this arrangement.

Typical implementation of a Fixed-bias, ultralinear Push-Pull output stage with a pair of pentode tubes (EL84 / 6BQ5)

Because the cathode is at (or very close to) ground potential, this requires the control grid to have a negative DC bias voltage applied to it, to regulate the flow of current through the tube.

If there was no negative bias applied to the tube, it would go into full conduction, the plate would glow red hot, fireworks will happen and that would be Bad, mmkay?
So we need to contrive to feed a constant negative voltage into the grid, along with the signal, to achieve the desired regulation.

The voltage required depends on several factors… as a very rough rule-of-thumb, take the screen voltage and divide it by the tube’s mu (gain) to get the maximum negative bias voltage likely to be needed

Looking at the circuit above… An EL84 has a mu of 20 and in this implementation the screen voltage is 350, which gives us 17.5 volts. Multiply by -1 since we’re dealing with negative volts. So we’re likely to need around -17.5 volts.In this case, our adjustment range is from -12.5 to -21.5 volts.

How to set up the bias adjustment resistor values
The bias voltage needs to be adjustable. Both tubes need to be drawing the same current, otherwise the net current through the transformer will not be zero, which will lead to magnetisation of the transformer core. This is a most undesirable situation and left unchecked, it will cause quantum fluctuations in the space-time continuum. Well ok maybe not that bad, but the transformer will saturate unevenly and distort the sound.

The usual approach is to use a voltage divider network with a potentiometer, as above.
Couple of points about this design.

  1. The more negative the voltage goes, the lower the current through the tube
  2. In this implementation, if the potentiometer fails, it will fail safe. The most common mode of failure with potentiometers is the wiper lifting off the track. If this happens, effectively the voltage at the grid of the tube will go full negative, reducing the current through the tube to (almost) zero. This is far more desirable than the voltage reaching zero and the tube immediately red-plating.
  3. Expanding on (2) – please don’t ever build this circuit with just the pot wiper connected to the grid. When the pot fails (and it will, eventually) it’ll likely take the tube with it.
  4. If you’re going to build this circuit, it’s intuitive to set it up so that clockwise rotation of the pot increases the current through the tube (ie brings the biasing voltage closer to zero)

The next question is – what value resistors will be needed? This is where some trial and error in the calculations is needed. Using Ohms’ law, these are the variables:

  • The desired bias voltage adjustment range
  • The input negative voltage from the power supply

From there, you can calculate the values for the resistors and potentiometer to give you the range you need.

This is where a spreadsheet can be mighty useful.

This is a screenshot from a spreadsheet I made that can help with this calculation. You can download this Excel worksheet here.

Put in the numbers in the red. Experiment with the values for R1 and R3 and the potentiometer, until you get the desired voltage range in the “Output V” column.

The “Build-Out R” represents the load seen by the preceding driver or phase-splitter stage, so watch the maximum “Rg-k” from the tube’s datasheet isn’t exceeded. In the case of an EL84, that value is 300K.

The columns of this spreadsheet:

  • Step – the setting on the potentiometer
  • Total R – the total resistance from the bias voltage to ground
  • Output V – the negative voltage as fed to the control grid
  • P(R1) – the amount of power dissipated by R1
  • P(R3) – power dissipated by R3
  • P(Pot) – the power dissipated by the potentiometer
  • Pot pwr % – the power dissipated by the pot expressed as a percentage of the pot’s total power rating AND the amount of track being used to conduct. 

About Pot pwr %
In the specifications for the potentiometer, there will be a power rating. However that power rating is across the entire length of the track. If the pot is set to half-position (assuming it’s a linear taper which in this design it is) then the power handling drops to half.So, this Pot pwr % column shows how much power the pot is dissipating as a percentage of its maximum considering the wiper position.

Using the circuit
Putting all this together, it’s easy to see how it works. Adjust the potentiometer for maximum negative voltage (wiper closest to the left, in this schematic). Power on the circuit and let it stabilize. Measure the voltage across the cathode resistor. Then adjust the potentiometer until the desired current is flowing through the tube.

What is the desired current through the tube?
Glad you asked. This depends on the tube itself, and the B+ voltage, and your preference regarding bias.As a rule of thumb, around 70% is the sweet spot with most types. If you bias low, that’s called “cold” biasing. The current through the tube is low, and the sound may take on a thin, glassy, brittle aesthetic. Also the distortion will increase.Bias too hot and you’ll shorten the life of the tubes.70% is the goldilocks zone.
So. Look at the tube’s datasheet. For an EL84 we see the maximum plate dissipation is 12 watts.70% of 12 watts is 8.4 wattsB+ is 350 volts so using Ohms’ law:

we solve for I at 24mA

Across 10 ohms (cathode resistor) our 24mA will give 0.24V

So we want to see 0.24V across the cathode resistor. Adjust the potentiometer until that’s the value shown. Then repeat for the other tube in the circuit, then do a final check that they’re both the same (or as close as you can get)

Please feel free to use my spreadsheet – I developed it to assist in choosing the resistor network values, and also to ensure the power rating of the potentiometer wasn’t being inadvertently exceeded at any setting.

A better time delay startup circuit

The amplifiers I’ve built so far have all incorporated a delayed turn-on circuit for the high voltage supply. The intention is to allow the 6 volt supply to turn on first and allow the valves to reach operating temperature before turning on the high voltage supply.

This is accomplished with a simple circuit based around a 555 timer IC in monostable mode, set up to a delay of around 25 seconds.

The circuit I’ve been using, while functioning, had a few problems. Driving a relay directly from the output of a 555 IC resulted in a lot of voltage drop through the IC and the relay coil voltage being low, for a 5V relay it was getting around 3.5 volts, fortunately this is still enough to trigger it, but less than ideal.

My re-design of the circuit was prompted by my addition of a 2-colour LED to the design, to glow red at initial turn-on but change to green when the timer activates and the HT voltage is turned on.

These LEDs are 2-pin, they work by reversing the polarity into them. So they’re 2 LEDs in one envelope, and depending on the polarity of the applied voltage, one will be forward biased and glowing, the other reverse biased and dark.

After breadboarding it and measuring carefully, this is the circuit I designed:

Note in this diagram my symbol library for the MOSFET is wrong… if you’re gonna use this same MOSFET be very aware its pinout (viewed from top) is S-G-D instead of G-D-S. So my pin numbers are wrong. Sorry about that.

The 7805 voltage regulator is not strictly necessary but it does result in a nice 4.9V across the relay coil.

The 330K and 68µF cap provide the time constant for the timer IC. The formula in this mode is:

T = 1.1 x R x C

The MOSFET Q1 buffers the output of the IC switching the negative on or off to the relay based on the voltage at the gate, which comes from the output of the IC at pin 3. This starts low until 25sec elapses then goes high and stays high until power down.

The two 330R resistors form a voltage divider, at the mid-point the voltage is 2.5V. When the relay is off, the + voltage will flow through the coil (which is around 62 Ohms) and then into the LED, then to ground through the lower 330R resistor. This results in a voltage drop of around 0.2 volts across the relay coil, not enough to turn it on.

When the IC turns on, the voltage appears at the gate of the MOSFET, switching the transistor on. This effectively shorts the Drain and Source, causing the negative to connect to the relay and the LED. At which point the return path for the LED is through the top 330R resistor, so this reverses the polarity across the LED causing it to change colour.

The reverse-biased diode across the relay is for flyback suppression.

After breadboarding, I’ve designed a single-layer PCB layout for this circuit which is 35mm x 35mm utilising a W02 rectifier.

On my board design I’ve also added a header for a regulated 5V power supply, in case it’s needed elsewhere (such as a tone control bypass relay for example).

The current and dissipation is such that no heatsinks are necessary on either the voltage regulator or MOSFET.

Be sure to put the relay on the AC side of the rectifier diodes, relays have a much easier time switching AC than DC and this is reflected in the voltage rating on the datasheet.